( ii ) a cos B cos C + b cos C cos A + c cos A cos B = Δ R

L.H.S. = a cos B · cos C + b cos C · cos A + c cos A · cos B = cos C a cos B + b cos C + c cos A · cos B = c cos C + c cos A · cos B = c cos π - A + B + cos A · cos B ∵ A + B + C = π = c - cos A + B + cos A · cos B = c - cos A · cos B + sin A · sin B + cos A · cos B = c sin A · sin B = 1 2 b c sin A 2 sin B b = △ b 2 sin B = △ R ∵ b sin B = 2 R

( iv ) cos 2 A 2 + cos 2 B 2 + cos 2 C 2 = 2 + r 2 R

Given that cos 2 A 2 + cos 2 B 2 + cos 2 C 2 We know that cos A = 2 cos 2 A 2 - 1 ⇒ cos 2 A 2 = 1 + cos A 2 = 1 + cos A 2 + 1 + cos B 2 + 1 + cos C 2 = 3 2 + 1 2 cos A + cos B + cos C . . . i cos A + cos B + cos C = cos A + 2 cos B + C 2 cos B - C 2 We know that in a triangle A + B + C = π = cos A + 2 sin A 2 cos B - C 2 ∵ cos B + C 2 = cos π 2 - A 2 We know that cos A = 1 - 2 sin 2 A 2 = 1 - 2 sin A 2 sin A 2 + cos B - C 2 = 1 - 2 sin A 2 cos B + C 2 - cos B - C 2 ∵ B + C 2 = π 2 - A 2 We know that cos B + C 2 - cos B - C 2 = - 2 sin B 2 sin C 2 = 1 + 4 sin A 2 sin B 2 sin C 2 . . . ii From equations i & ii , cos 2 A 2 + cos 2 B 2 + cos 2 C 2 = 3 2 + 1 2 1 + 4 sin A 2 sin B 2 sin C 2 We know that r = 4 R sin A 2 sin B 2 sin C 2 = 3 2 + 1 2 + r 2 R = 2 + r 2 R Hence, proved.

( iii ) 1 r 2 + 1 r 1 2 + 1 r 2 2 + 1 r 3 2 = a 2 + b 2 + c 2 Δ 2

Given that L.H.S. = 1 r 2 + 1 r 1 2 + 1 r 2 2 + 1 r 3 2 We know that r = ∆ s , r 1 = ∆ s - a , r 2 = ∆ s - b , r 3 = ∆ s - c . Here ∆ , s are area and semi perimeter of triangle respectively. = s 2 ∆ 2 + s - a 2 ∆ 2 + s - b 2 ∆ 2 + s - c 2 ∆ 2 = 1 ∆ 2 s 2 + s 2 + a 2 - 2 a s + s 2 + b 2 - 2 b s + s 2 + c 2 - 2 c s = 1 ∆ 2 4 s 2 + a 2 + b 2 + c 2 - 2 s a + b + c = 1 ∆ 2 4 s 2 + a 2 + b 2 + c 2 - 2 s 2 s ∵ 2 s = a + b + c = 1 ∆ 2 4 s 2 + a 2 + b 2 + c 2 - 4 s 2 = 1 ∆ 2 a 2 + b 2 + c 2 = R.H.S.

( iv ) 1 r + 1 r 1 + 1 r 2 + 1 r 3 2 = 4 r 1 r 1 + 1 r 2 + 1 r 3

We know that r = ∆ s , r 1 = ∆ s - a , r 2 = ∆ s - b , r 3 = ∆ s - c 1 r + 1 r 1 + 1 r 2 + 1 r 3 2 = s ∆ + s - a ∆ + s - b ∆ + s - c ∆ 2 = 1 ∆ 2 4 s - a + b + c 2 = 1 ∆ 2 2 s 2 = 4 s 2 ∆ 2 . . . i 4 r 1 r 1 + 1 r 2 + 1 r 3 = 4 s ∆ s - a ∆ + s - b ∆ + s - c ∆ = 4 s ∆ 2 3 s - a + b + c = 4 s ∆ 2 3 s - 2 s = 4 s 2 ∆ 2 . . . ii From equations i & ii , ⇒ 1 r + 1 r 1 + 1 r 2 + 1 r 3 2 = 4 r 1 r 1 + 1 r 2 + 1 r 3

( v ) b c - r 2 r 3 r 1 = c a - r 3 r 1 r 2 = a b - r 1 r 2 r 3 = r

Given that a b - r 1 r 2 r 3 We know that a sin A = b sin B = c sin C = 2 R ⇒ a = 2 R sin A , b = 2 R sin B , c = 2 R sin C . We know that r 1 = 4 R sin A 2 cos B 2 cos C 2 , r 2 = 4 R cos A 2 sin B 2 cos C 2 , r 3 = 4 R cos A 2 cos B 2 sin C 2 = 2 R sin A 2 R sin B - 4 R sin A 2 cos B 2 cos C 2 4 R cos A 2 sin B 2 cos C 2 4 R cos A 2 cos B 2 sin C 2 = 4 R 2 sin A sin B - 4 R 2 cos 2 C 2 2 sin A 2 cos A 2 2 sin B 2 cos B 2 4 R cos A 2 cos B 2 sin C 2 We know that sin 2 x = 2 sin x cos x = 4 R 2 sin A sin B 1 - cos 2 C 2 4 R cos A 2 cos B 2 sin C 2 = R · 2 sin A 2 cos A 2 · 2 sin B 2 cos B 2 · sin 2 C 2 cos A 2 cos B 2 sin C 2 = 4 R sin A 2 sin B 2 sin C 2 = r Similarly, b c - r 2 r 3 r 1 = c a - r 1 r 3 r 2 = r ∴ b c - r 2 r 3 r 1 = c a - r 1 r 3 r 2 = a b - r 1 r 2 r 3 = r

If α , β , γ are the respective altitudes of a triangle A B C , prove that 1 α + 1 β - 1 γ = 2 a b ( a + b + c ) Δ cos 2 C 2

As we know that, area of triangle = 1 2 × base × height So, ∆ = 1 2 × a × α = 1 2 × b × β = 1 2 × c × γ ⇒ α = 2 ∆ a , β = 2 ∆ b , γ = 2 ∆ c Now, LHS = 1 α + 1 β - 1 γ = a + b - c 2 ∆ As we know that, 2 s = a + b + c LHS = 2 ( s - c ) ∆ RHS = 2 a b ( a + b + c ) ∆ cos 2 C 2 As we know that, 2 s = a + b + c , cos 2 C 2 = s ( s - c ) a b RHS = 2 a b 2 s ∆ × s ( s - c ) a b ⇒ LHS = RHS Hence, proved.

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 33 - Properties of Triangle>Exercise-1

Embibe Experts Solutions for Chapter: Properties of Triangle, Exercise 1: Exercise-1

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Properties of Triangle, Exercise 1: Exercise-1

Attempt the free practice questions on Chapter 33: Properties of Triangle, Exercise 1: Exercise-1 with hints and solutions to strengthen your understanding. Alpha Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Properties of Triangle, Exercise 1: Exercise-1 with Hints & Solutions

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 33 - Properties of Triangle>Exercise-1>Q 27

$(ii)$ $a \cos B \cos C + b \cos C \cos A + c \cos A \cos B = \frac{Δ}{R}$

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JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 33 - Properties of Triangle>Exercise-1>Q 28

$(iv)$ $\cos^{2} \frac{A}{2} + \cos^{2} \frac{B}{2} + \cos^{2} \frac{C}{2} = 2 + \frac{r}{2 R}$

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JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 33 - Properties of Triangle>Exercise-1>Q 29

$(v)$ $a \cot A + b \cot B + c \cot C = 2 (R + r)$

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JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 33 - Properties of Triangle>Exercise-1>Q 31

$(ii)$ $r_{1} + r_{2} - r_{3} + r = 4 R \cos C$

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 33 - Properties of Triangle>Exercise-1>Q 32

$(iii)$ $\frac{1}{r^{2}} + \frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}} = \frac{a^{2} + b^{2} + c^{2}}{Δ^{2}}$

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 33 - Properties of Triangle>Exercise-1>Q 33

$(iv)$ ${(\frac{1}{r} + \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}})}^{2} = \frac{4}{r} (\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}})$